Fiber Optic Transmission

SL/HL – Option F.3

Reflection/Refraction

• Reflection– A wave encounters a boundary between two

mediums and cannot pass through– The angle of incidence is always equal to the

angle of reflection

• Refraction– When a wave passes through boundary into a

new medium its speed changes– The wave will change directions based on the

change in its speed

Refraction

• If the wave speeds up it will bend away from the normal line

• If the wave slows down it will bend towards the normal line

• Snell’s Law

sinθi cr = sinθr ci

sin sini i r rn n

Critical Angle

• When traveling into air from some medium, light will always speed up, thus increasing the angle– If it speeds up enough, the angle of refraction will

be 90 degrees• This means that the refracted ray will travel along the

edge of the boundary

Critical Angle

• When light strikes the boundary at the critical angle or greater, the wave is totally reflected back into the first medium

• Here n1 is the index of refraction of the medium the light starts in

Total Internal Reflection

• Usually when a wave reaches a boundary between mediums it is partially reflected and partially refracted– When the critical angle

is exceeded the entire wave is reflected back within the medium

– The wave doesn’t lose any energy

Optical Fiber

• Fiber optic cable is made of thin, clear glass or plastic

• Once light enters the cable it is totally internally reflected until it reaches the far end– Actual optical fiber is step indexed

• There is another layer between the core and the outside

• This is so the fibers can be bundled together

Critical Angle

• The critical angle is that angle of incidence for which the angle of refraction is 90 degrees.

• n1 sinθc = n2 sin90• Θc = sin-1 (n1/n2)

Practice Problem

• A ray of light in water (refractive index 1.33) is incident from water on a water-air boundary.

• Calculate the critical angle of the water-air boundary.

Practice Problem

• A ray is emitted at point P is internally reflected. What can you deduce about the refractive index of the liquid?

Air = 1.00

Liquid = ?

8cm

12cm

Dispersion

• Modal– Not all the waves that enter make it to the other end, only

certain ones– The possible paths are called modes

• Material– Because different frequencies have different refractive

indices, they have different paths• These can both cause problems if the bits of data arrive out of order • More direct modes are faster• Laser light and single mode cable reduce

these effects

Acceptance Angle

• The maximum angle of incidence a ray entering a fibre resulting in total internal reflection is called the acceptance angle.

• A = sin-1 √(n12 – n2

2)

Practice Problem

• The refractive index of the core of an optical fibre is 1.50 and that of the cladding is 1.40. Calculate the acceptance angle of the fibre.

Material Dispersion

Attenuation

• Attenuation is the opposite of amplification– As a signal travels through a cable it will slowly

lose intensity as energy is lost– Attenuation is measured in decibels (dB)

• The 10 at the beginning is to convert to decibels• Generally measured in dBkm-1

10 10

10 log ( ) 10log ( )

i

o

PPower inattenuation

Power out P

Amplifiers

• Even with reshaping, signals still attenuate over the length of the cable– Amplifiers along the cable increase the signal

strength to keep it going

• The same equation for Attenuation applies to Amplifiers

Practice Problem

• An amplifier amplifies an incoming signal of power 0.34mW to a signal of power 2.2mW. Calculate the power gain of the amplifier in decibels.

Practice Problem

• A signal starting with 10mW of power is reduced by 19 decibels over 50km. Calculate the power lost per km.

Reshapers

• Monomode fibers can eliminate modal dispersion and lasers cut down on material dispersion, but it is not completely eliminated– Over a long distance individual pulses can start to

overlap each other– Every 40-60km is a reshaper which will detect and

reshape the signal• Has its own laser which sends a ‘new’ signal

Noise

• One advantage to using fiber optics is that it is not particularly susceptible to noise– Any noise that does occur is generally due to

random light entering the end of the cable– The power ratio of noise to signal in fiber optics is

generally in the range of 10-17 or 10-18

10 10

10 log ( ) 10log ( )

i

o

PPower inattenuation

Power out P Signal to

Noise (dB)

Practice Problem

• The minimum SNR considered acceptable for a certain signal is 30dB. If the power of the noise is 2.0mW, calculate the least acceptable signal power.